Answer
$x_1=\frac{3}{2}$
$x_2=0$
$x_3=\frac{-1}{2}$
Work Step by Step
We are given:
$3x_1-2x_2+x_3=4$
$x_1+x_2+x_3=2$
$x_1+x_3=1$
which can also be written as:
$A=\begin{bmatrix}
3 & -2&1\\
1 & 1&-1\\
1&0&1
\end{bmatrix} \rightarrow \det(A)=3+4-1=6$
$B_1=\begin{bmatrix}
4 &-2&1\\
2 & 1&-1\\
1&0&1
\end{bmatrix} \rightarrow \det(A)=4+6-1=9$
$B_2=\begin{bmatrix}
3 &4&1\\
1& 2&-1\\
1&1&1
\end{bmatrix} \rightarrow \det(A)=9-8-1=0$
$B_3=\begin{bmatrix}
3&-2&4\\
1& 1&2\\
1&0&1
\end{bmatrix} \rightarrow \det(A)=3-2-4=-3$
Use Cramer’s rule: $x_k=\frac{\det(B_k)}{\det(A)}$ to find the results:
$x_1=\frac{\det(B_1)}{\det(A)}=\frac{9}{6}=\frac{3}{2}$
$x_2=\frac{\det(B_2)}{\det(A)}=\frac{0}{6}=0$
$x_3=\frac{\det(B_3)}{\det(A)}=\frac{-3}{6}=\frac{-1}{2}$
